Problem: Solve for $x$ and $y$ using elimination. $\begin{align*}-6x+6y &= -8 \\ x+y &= 2\end{align*}$
Solution: We can eliminate $y$ when its corresponding coefficients are negative inverses. Recalling our knowledge of least common multiples, multiply the top equation by $-1$ and the bottom equation by $6$ $\begin{align*}6x-6y &= 8\\ 6x+6y &= 12\end{align*}$ Add the top and bottom equations. $12x = 20$ Divide both sides by $12$ and reduce as necessary. $x = \dfrac{5}{3}$ Substitute $\dfrac{5}{3}$ for $x$ in the top equation. $-6( \dfrac{5}{3})+6y = -8$ $-10+6y = -8$ $6y = 2$ $y = \dfrac{1}{3}$ The solution is $\enspace x = \dfrac{5}{3}, \enspace y = \dfrac{1}{3}$.